Let $f(x)=\dfrac{\ln(x)}{x^2}$. What is the absolute maximum value of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{e}$ (Choice B) B $\dfrac{1}{e^2}$ (Choice C) C $\dfrac{1}{2e}$ (Choice D) D $f$ has no maximum value
Answer: Let's first find the relative extremum points of $f$, and then consider them along with the function's end behavior in both directions. Note that the domain of $f$ is all real numbers such that $x>0$. We start with finding the critical points of $f$. The derivative of $f$ is $f'(x)=\dfrac{1-2\ln(x)}{x^3}$. $f'(x)=0$ for $x=\sqrt e$. $f'$ is defined for all real numbers in the domain of $f$. Therefore, our only critical point is $x=\sqrt e$. Our critical point divides the function's domain (which is all real numbers such that $x>0$ ) into two intervals: $0$ $1$ $2$ $3$ $4$ $(0,\sqrt e)$ $(\sqrt e,\infty)$ $\sqrt e$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $(0, \sqrt e)$ $x=\dfrac12$ $f'(\dfrac12)=8+16\ln(2)>0$ $f$ is increasing $\nearrow$ $(\sqrt e,\infty)$ $x=2$ $f'(2)=\dfrac{1-\ln(4)}{8}<0$ $f$ is decreasing $\searrow$ Let's imagine ourselves walking on the graph of $f$, starting at the left end of the domain $($ which is $0$ $)$ and going all the way to the right (until $+\infty$ ). According to the table, we will start by going up until we reach $x=\sqrt e$. Then, we will be forever going down. Therefore, $f$ must obtain its absolute maximum value at $x=\sqrt e$. We are asked to find that maximum value, which is $f(\sqrt e)=\dfrac{1}{2e}$. In conclusion, the absolute maximum value of $f$ is $\dfrac{1}{2e}$.